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kodGreya [7K]
2 years ago
13

What is the percentage of carbon in 5.00 g of sucrose

Chemistry
1 answer:
Irina-Kira [14]2 years ago
3 0
The answer is rounded off to 0.0278 mol C6H12O6
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What is a membrane?
9966 [12]
B I think, it has layers that protects the cells nucleus
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3 years ago
A catalyst lowers the activation energy for both the forward and the reverse reactions in an equilibrium system, so it has no ef
rosijanka [135]

Answer:

True.

Explanation:

  • Catalyst increases the rate of the reaction without affecting the equilibrium position.
  • Catalyst increases the rate via lowering the activation energy of the reaction.
  • This can occur via passing the reaction in alternative pathway (changing the mechanism).
  • The activation energy is the difference in potential energies between the reactants and transition state (for the forward reaction) and it is the difference in potential energies between the products and transition state (for the reverse reaction).
  • in the presence of a catalyst, the activation energy is lowered by lowering the energy of the transition state, which is the rate-determining step, catalysts reduce the required energy of activation to allow a reaction to proceed and, in the case of a reversible reaction, reach equilibrium more rapidly.
  • with adding a catalyst, both the forward and reverse reaction rates will speed up equally, which allowing the system to reach equilibrium faster.
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5 0
3 years ago
Read 2 more answers
PLEASE HELP ME!! DIMENSIONAL ANALYSIS<br> CHEMISTRY<br> DUE IN 5 MINS!!!
Eddi Din [679]

Answer:

12 mi/h

Explanation:

Step 1: Given data

  • Total distance (d): 6 km
  • Time elapsed (t): 19 min

Step 2: Convert "d" to miles

We will use the conversion factor 1 mi = 1.60934 km.

6 km × 1 mi/1.60934 km = 3.7 mi

Step 3: Convert "t" to hours

We will use the conversion factor 1 h = 60 min.

19 min × 1 h/60 min = 0.32 h

Step 4: Calculate the average speed of the runner (s)

The speed is equal to the quotient between the total distance and the time elapsed.

s = d/t

s = 3.7 mi/0.32 h = 12 mi/h

6 0
3 years ago
A student wants to reclaim the iron from an 18.0-gram sample of iron(III) oxide, which
lilavasa [31]

Answer:

m_{Fe}=12.6gFe

Explanation:

Hello,

In this case, since we have grams of iron (III) oxide whose molar mass is 159.69 g/mol are able to compute the produced grams of iron by using its atomic mass that is 55.845 g/mol and their 2:4 molar ratio in the chemical reaction:

m_{Fe}=18.0gFe_2O_3*\frac{1molFe_2O_3}{159.69gFe_2O_3}*\frac{4molFe_2O_3}{2molFe_2O_3} *\frac{55.845gFe}{1molFe_2O_3} \\\\m_{Fe}=12.6gFe

Best regards.

3 0
2 years ago
Read 2 more answers
What is the empirical formula for a compound if 300.00 g of it is known to contain 82.46224 g of molybdenum, 45.741 g of chlorin
Marysya12 [62]

Answer:

MoClBr₂

Explanation:

First we calculate the mass of bromine in the compound:

  • 300.00 g - (82.46224 g + 45.741 g) = 171.79676 g

Then we<u> calculate the number of moles of each element</u>, using their <em>respective molar masses</em>:

  • 82.46224 g Mo ÷ 95.95 g/mol = 0.9594 mol Mo
  • 45.741 g Cl ÷ 35.45 g/mol = 1.290 mol Cl
  • 171.79676 g Br ÷79.9 g/mol = 2.150 mol Br

Now we <u>divide those numbers of moles by the lowest number among them</u>:

  • 0.9594 mol Mo / 0.9594 = 1
  • 1.290 mol Cl / 0.9594 = 1.34 ≅ 1
  • 2.150 mol Br / 0.9594 = 2.24 ≅ 2

Meaning the empirical formula is MoClBr₂.

8 0
3 years ago
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